What are the solutions to the equation r^2+12r+17=0 ?

The solutions to the equation r^2+12r+17=0 are r=-6+sqrt(19),r=-6-sqrt(19)

Find the zeros of r^2+12r+17=0

The zeros of r^2+12r+17=0 are r=-6+sqrt(19),r=-6-sqrt(19)

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r^{2} + 12 r + 17 = 0

r = - 6 + 19, r = - 6 - 19

Decimal

r = - 1.64110 \dots, r = - 10.35889 \dots

Solution steps

r^{2} + 12 r + 17 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 17}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 17 = 219

r_{1, 2} = \frac{- 12 \pm 2 19}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 12 + 2 19}{2 \cdot 1}, r_{2} = \frac{- 12 - 2 19}{2 \cdot 1}

r = \frac{- 12 + 2 19}{2 \cdot 1} : - 6 + 19

r = \frac{- 12 - 2 19}{2 \cdot 1} : - 6 - 19

The solutions to the quadratic equation are:

r = - 6 + 19, r = - 6 - 19

x^{2} - x + 4

9 x^{2} - 15 x - 6 = 0

(40 - x) (60 - x) = 800

2 x^{2} + 160 = 36 x

y, y^{2} + 2 y + 1 = 0

What are the solutions to the equation r^2+12r+17=0 ?
The solutions to the equation r^2+12r+17=0 are r=-6+sqrt(19),r=-6-sqrt(19)
Find the zeros of r^2+12r+17=0
The zeros of r^2+12r+17=0 are r=-6+sqrt(19),r=-6-sqrt(19)